Physics >> Electrostatics >> AC Circuit
A.C. Circuit
Let a pure resistance R a pure inductance L and an ideal capacitor of capacitance C be connected in series to a source of alternating fig. As R, L, C is in series therefore, current at any instant through the three elements has the same amplitude and phase let it be represented by
I = I0 sin wt
However a voltage across each element bears a different phase relationship with the current. Now,
The maximum voltage across R is
V R = I0 R
In fig. current phasor Io is represented along OX.
As VR is in phase with current it is represented by the vector OA, along OX.
The maximum voltage across L is
VL = I0 XL
As voltage across the inductor leads the current by 90° it is represented by OB along OY, 90° ahead of Io.
The maximum voltage across C is
Vc = I0 Xc
As voltage across the capacitor lags behind the alternating current by 90° it is represented by OC rotated clocking through 90° from the direction of Io OC is along OY.
As the voltages across L and C have a phase difference of 180° the net reactive voltage is (VL – VC), assuming that VL > VC.
In fig. it is represented by OB the resultant of OA and OB is the diagonal OK of the rectangle OAKB. Hence the vector sum of VR VL and VC is phasor Eo represented by OK, making an angle ∅ with current phasor Io.
As OK = √OA2 + OB2
∴ E0 = √V2R + (VL – Vc)2 = √((I0) R)2 + (I0 XL – I0 XC)2
E0 = I0 √R2 + (XL – XC) 2
The total effective resistance of RLC circuit is called impedance of the circuit. It is represented by Z, where
Z = E0 / I0 = √R2 + (XL – XC)2
From it is clear that in an a.c circuit containing R, L, C the voltage leads the current by a phases angle ∅ where
Tan ∅ = (AK/OA) = (OB/OA) = (VL – VC)/VR
= (I0 XL – I0 Xc)/I0 R or
Tan ∅ = (XL – XC)/R
∴ The alternating in the RLC circuit would be represented by
E = E0 sin (wt + ∅)
Three cases arise:
When XL = Yc tan ∅ = 0 ∴ ∅ = 0
Hence voltage and current are in the same phase. The a.c. circuit is non- inductive.
When XL > Xc. Tan ∅ is positive. Therefore ∅ is positive hence voltage leads the current by a phase angle ∅. The a.c circuit is inductance dominated circuit.
When X L < Xc tan ∅ is negative. Therefore ∅ is negative angle ∅. The circuit is capacitance dominated circuit.
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